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Magazine Chapter 14: Problem 24
Solve the following relations for \(x\) and \(y,\) and compute the Jacobian \(J(u, v)\). $$u=x y, v=x$$
Short Answer
Expert verified
Answer: The Jacobian of the transformation is given by J(u, v) = -\(\frac{1}{v}\).
Step by step solution
01
Solve for x and y in terms of u and v
The given relations are u = xy and v = x. Solve the second relation for x to get x = v, and then substitute this into the first relation to solve for y in terms of u and v:
$$u = x y = v y \implies y = \frac{u}{v}$$
Now, x and y have been expressed in terms of u and v. x = v and y = \(\frac{u}{v}\).
02
Calculate partial derivatives of x and y with respect to u and v
Next, we need to calculate the following partial derivatives:
- \(\frac{\partial x}{\partial u}\)
- \(\frac{\partial x}{\partial v}\)
- \(\frac{\partial y}{\partial u}\)
- \(\frac{\partial y}{\partial v}\)
Using these results, we can get:
- \(\frac{\partial x}{\partial u} = 0\) (x doesn't depend on u)
- \(\frac{\partial x}{\partial v} = 1\) (x is equal to v, so it has a constant derivative with respect to v)
- \(\frac{\partial y}{\partial u} = \frac{1}{v}\) (y is equal to \(\frac{u}{v}\), so its derivative with respect to u is \(\frac{1}{v}\))
- \(\frac{\partial y}{\partial v} = - \frac{u}{v^2}\) (y is equal to \(\frac{u}{v}\), so its derivative with respect to v is -\(\frac{u}{v^2}\))
03
Form the Jacobian matrix and compute J(u, v)
Now that we have the partial derivatives, we can form the Jacobian matrix as follows:
$$J(u,v) =
\begin{bmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{1}{v} & -\frac{u}{v^2}
\end{bmatrix}$$
Finally, compute the determinant of the Jacobian matrix:
$$J(u,v) = (0) (-\frac{u}{v^2}) - (1)(\frac{1}{v}) = -\frac{1}{v}$$
The Jacobian J(u, v) is equal to \(-\frac{1}{v}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives help us understand how a function changes as one of its variables changes, while keeping the others constant. In this exercise, we calculate the partial derivatives of functions with respect to given variables.
- When solving the relation, we express variables in terms of others, like \( x = v \) and \( y = \frac{u}{v} \).
- Calculating partial derivatives involves differentiating a variable concerning others. For instance, \( \frac{\partial x}{\partial u} = 0 \) since \( x \) is not dependent on \( u \).
- For \( y = \frac{u}{v} \), its derivative with respect to \( u \) is \( \frac{1}{v} \), because changing \( u \) adjusts \( y \) in a proportional manner.
- On the other hand, the derivative of \( y \) with respect to \( v \) turns out to be \(-\frac{u}{v^2} \), illustrating a more complex relationship involving division.
Jacobian Matrix
A Jacobian matrix brings together the partial derivatives of multiple variables to form a cohesive structure. It's especially essential in multi-variable calculus when transforming from one coordinate space to another.
- The Jacobian matrix outlines how each variable鈥攕uch as \( x \) and \( y \)鈥攃hanges in relation to input variables, denoted here by \( u \) and \( v \).
- For this exercise, the matrix is constructed by placing each partial derivative in a specific position. Example entries include \( \frac{\partial x}{\partial u} \) and \( \frac{\partial y}{\partial v} \).
- We then form the matrix:\[J(u,v) = \begin{bmatrix}0 & 1 \\frac{1}{v} & -\frac{u}{v^2}\end{bmatrix}\]
Determinant of a Matrix
The determinant of a matrix provides a scalar value that offers insight into various properties like invertibility or volume scaling in transformations. Calculating the determinant is straightforward yet significant:
- For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is \( ad - bc \).
- In this exercise, our Jacobian matrix has the form: \( \begin{bmatrix} 0 & 1 \ \frac{1}{v} & -\frac{u}{v^2} \end{bmatrix} \).
- Computing the determinant involves multiplying elements (0 by \(-\frac{u}{v^2}\)) and subtracting (1 by \(\frac{1}{v}\)), thus resulting in \(-\frac{1}{v}\).
Functions of Multiple Variables
Functions of multiple variables are common in higher math, involving several inputs to produce a single output. These functions are essential for understanding complex systems:
- In our exercise, variables \(x\) and \(y\) depend on inputs \(u\) and \(v\). Specifically, \(u = x y\) and \(v = x\), displaying a multi-variable relationship.
- Often, these types of functions require solving simultaneous equations to express one set of variables in terms of another.
- Understanding the behavior of these functions, through tools like partial derivatives and Jacobian matrices, helps in fields like physics and engineering where systems depend on several parameters.
